is battery drain linear with resect to Mah?

[J M Fahey]

Then i tried Juan's dying battery idea but i don't understand what you mean by "reaching bass".

We are talking feeding an active Bass, arenīt we?
Reaching Bass means what voltage is actually reaching the Bass PCB.
IF you are instead feeding some pedal ???????? then feed it *through* the resistor I mentioned.

NO WAY any PCB, active bass or effects pedal gets *only* 0.3V less than what battery or supply offers, NO WAY.
Even less *through* 100k.

You are either wiring those resistors in parallel with battery, or still measuring at the battery end, of course you need to measure at the load side, not at the battery/supply side. .

There is a reason such resistors are called "voltage dropper"

In any case i tried 1k in series with the + then gradually more resistance with no change. So i jumped all the way to 100k and only .3v less ! 9.7 vs the 10.05v where the battery is after a bit of use. So not sure what i am doing wrong

[daz]

Did you compare loaded battery voltages (i.e. in circuit with effects engaged) or with batteries disconnected?

yeah, both but theres never much difference, usually on the order of .01-.03v and the alkaline about .1 or .2 different.

[J M Fahey]

Check what dropping resistors R66-67 do, when going from +40V supply towards Preamp load [A]

[daz]

We are talking feeding an active Bass, arenīt we?
Reaching Bass means what voltage is actually reaching the Bass PCB.
IF you are instead feeding some pedal ???????? then feed it *through* the resistor I mentioned.

NO WAY any PCB, active bass or effects pedal gets *only* 0.3V less than what battery or supply offers, NO WAY.
Even less *through* 100k.

You are either wiring those resistors in parallel with battery, or still measuring at the battery end, of course you need to measure at the load side, not at the battery/supply side. .

There is a reason such resistors are called "voltage dropper"

Are u talking with the battery connected to the pedals or just measuring the battery with a resistor? Because i just checked again with a 100k resistor with the battery NOT connected to the pedals and it is almost the same voltage with or w/o the resistor. I took pics in case thats why you meant, but before i go to the trouble of uploading them let me know if u r talking about when connected to the pedals and whether that could make such a huge difference. But battery with 100k in series with the + and measuring from the other end of the resistor, NOT the battery side, i get a little over .1v less, something like that. I have pics showing both ways with meter setting and readout and battery and resistor all in frame and it shows almost the same with meter hot lead at either end of the resistor. If that should be far greater let me know and i will upload them so you believe me. If you mean when connected to the pedal board let me know and i will try that method.

[daz]

Ok, i see....u meant under load. I tried it and 100k makes 5v. I would think much lower from what u said but in any case i get it now. Oddly the pedals seem to work fine at 5 v ! (tho only a minute trying it)

[daz]

Oh, and no, not a bass....a guitar pedalboard

[J M Fahey]

Ok, i see....u meant under load. I tried it and 100k makes 5v. I would think much lower from what u said but in any case i get it now. Oddly the pedals seem to work fine at 5 v ! (tho only a minute trying it)

I never suggested 100k but 500 ohm to 1k for starters and then fine tune.

YOU mentioned 100k and I said NO WAY.
In fact I am surprised you get 5V at the board through 100k, 5V (10V-5V=5V) though 100k means board is eating incredible/impossible 0.05mA=50uA
No pedal eats that little.

And of course itīs measured feeding the preamp, board, whatever.

And again, I care what voltage reaches the board, not whatīs measured at the battery end.

Not sure we arenīt having a language problem here , my instructions are quite clear and simple.

[daz]

Well yeah, but what threw me off is "reaching bass". I thought it was some term/phrase i wasn't familiar with. I needed something like "read it under load" or such. I never mentioned anything about using it in an active bass, in fact my initial post describes using it with pedals. So yeah, there definitely was a communication issue. I also never suggested u told me to try a 100k. I tried that because after trying your 1k suggestion i kept adding more resistance and because i wasn't testing it under load the voltage was barely changing. So i radically upped the resistance to 100k just to see how much it would take to lower the voltage.

[J M Fahey]

Cool.
Just test it under the real load, the board itself.

Canīt suggest a specific value because I donīt know the current it eats, but try different values until you get 8.4 V **at the board** or whatever makes it (you ) happy.

Battery will need to be "outside" on its own battery clip to be able to insert the series voltage dropping resistor or you will need to cut the "+9V" track from clip to +9V rail to insert resistor if you want to use some board built in battery holder.

[daz]

I did and got it down to mid 8v. Not sure if my ears are burned out from all the listening and trying different things or what but it seems less of an issue now even plugged straight in w/o a resistor. I wonder if the battery break in is to blame. The instructions suggest they need 3-5 cycles to "reach peak performance" which may be akin to them breaking in in some way. Says to use them completely then charge them 3-5 time so i am leaving them in the board and the switch (put a switch in to lift ground when not using) on till fully drained then charging them 3 times. Or maybe because it's down to a tad under 10v now. Not sure if i will bother inserting a resistor but i will wait till i have fresh ears and have cycled the batteries 3 times. As to the logistics of how to insert a resistor, thats not a problem. If i'm good at anything it's the mechanical end of things.

[daz]

Cool.
Just test it under the real load, the board itself.

Canīt suggest a specific value because I donīt know the current it eats, but try different values until you get 8.4 V **at the board** or whatever makes it (you ) happy.

Battery will need to be "outside" on its own battery clip to be able to insert the series voltage dropping resistor or you will need to cut the "+9V" track from clip to +9V rail to insert resistor if you want to use some board built in battery holder.

Here's a different question Juan. What about using a regulator to bring a freshly charged battery to say 8.5v so that rather than starting with about 10v and having to wait for it to drain to 8.5 then only having a short time at that voltage, it would stay at 8.5v for almost the entire life cycle? Is that possible with a regulator and if so can you point me to a simple schematic ? Because using resistors will get me to the 8.5v right away but then it will drop too low for a short time and die quickly as you (or someone) said rechargeables do.

[daz]

Maybe i should clarify. I know how i can get 9v or 5 volts with a regulator....just get one of that value. But if i want 8 or 8.5v i can't use a resistor because it will cause the same issue as using a resistor to drop voltage which is that it will not stay at the regulator's output voltage as it drains. The whole reason i was thinking about using a regulator is so it would stay at that voltage all the way from a fresh charge to say 8v and only change once it goes below. Hope thats clear.

[Enzo]

I have 8v regulators in stock, and if you want 8.5v, just put a 1N4148 or something in its ground leg to lift it. One issue with regulators is the dropout voltage. In other words its input has to be somewhat higher than the output. An 8v regulator will not keep the output up to 8v if the input is down to 8.1 or some such.

[Mick Bailey]

Look for an LDO (Low Drop Out) regulator. These can have a dropout voltage of 100mv or less, rather than the 2v to 3.5v of a typical linear regulator. Also bear in mind that a regulator uses power in its own right and many LDO regulators feature very low current consumption - sometimes just a few micro amps. This is an important consideration with battery powered equipment.

[daz]

Look for an LDO (Low Drop Out) regulator. These can have a dropout voltage of 100mv or less, rather than the 2v to 3.5v of a typical linear regulator. Also bear in mind that a regulator uses power in its own right and many LDO regulators feature very low current consumption - sometimes just a few micro amps. This is an important consideration with battery powered equipment.

Ok, bear with me because i'm not sure i understand why to look for ope of those. If the typical one doesn't dropout till 3.5v, why isn't that fine for my purpose? I'm not going to continue using a batter when the voltage gets anywhere near that low. So before i look for one i need to know why that would be necassary at all. I think a 8v as Enzo suggested would be fine, as these start fully charged at 10 and pretty much drop rapidly to needing a recharge by the time the are i think somewhere around 7 or so. So 8v regulator would be fine and should keep it at 8v most of it;s charge cycle which is fine, or like Enzo said use a diode on the -.